The effect of gravity on a speeding bullet

Discussion in 'Physics & Math' started by Magical Realist, Jan 31, 2017.

  1. RajeshTrivedi Valued Senior Member

    And now you wish to bring in engine thrust and all? I am sticking to OP, there is no generalization. Bullet shot in downwards direction from a flying aircraft.

    You are bringing all these additional stuff that under a special aerodynamics even if horizontal speed is non zero the vertical terminal speed is achieved. You know thrust can create any kind of system, I can make an object fly in any direction at constant speed by sufficiently controlled thrust. Are we discussing that??
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  3. Baldeee Valued Senior Member

    Yes, the drag acts opposite the direction of travel, and yes, the vertical component of drag is not generally proportional to the vertical velocity.
    This is relatively simple, and understood.
    My point was that the magnitude of the drag is the sum of the magnitudes of the drags of the horizontal and vertical components of the velocity.
    That is how I interpreted RT's comment:
    By way of illustration, take an object travelling 8 m/s vertically and 2 m/s horizontally.
    The resultant velocity is sqrt(68) m/s along the direction 2i+8j.
    The resultant drag is 68x where x is the various other components of drag, the p, the S, the Cd etc.
    Now the magnitude of the drag due to horizontal velocity is 4x, and due to vertical velocity is 64x.
    The sum is 68x.
    Thus F(824) is the same magnitude as F(2)i + F(8)j., contrary to what RT was claiming.

    If I have misunderstood hat RT was claiming, so be it, but I responded as I have interpreted it, and I stand by what I have said.

    Further, you seem to be slipping between stating drag being proportional to speed and being proportional to the square of the speed.
    But no issue as I know what you intended.
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  5. Baldeee Valued Senior Member

    It's all well and good saying "don't let it confuse you" but if what you write is confusing and difficult to interpret, then such warnings are moot.
    Yes, the drag coefficient can be speed dependent.
    So what?
    When you are looking at the instantaneous forces etc, as my comments were doing, variability of the coefficient over time is irrelevant.
    And yes, cross section is different, but not when you are working out the magnitudes of horizontal and vertical drag.
    For that you use the cross section and drag coefficient in the actual direction of travel, so are taken to be the same.
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  7. Sarkus Hippomonstrosesquippedalo phobe Valued Senior Member

    I was simply using the example of aircraft thrust to explain a point, about how all one needs to achieve is a balance of forces. Do you deny that that will achieve terminal velocity? As such it is entirely relevant to the OP, and your dismissal of it is nothing but flagrant dishonesty.
    It is not necessarily additional. That is simply your assumption. Do you deny that to achieve terminal velocity one must simply balance all the forces at play? Do you deny that this can result in forward motion as well as vertical?
    Stop arguing strawmen, please. The issue of thrust was to make the point that all one needs to achieve is a balance of forces. Are you denying that that is what is required to achieve terminal velocity? Do you deny that in balancing the forces - without thrust (just so as to be perfectly clear) - one can end with a resultant motion that has both a horizontal and vertical component?

    Go on, bleat about how you're only sticking with the OP. But if you can't even acknowledge the basic principles at play then why should anyone take what you say with any seriousness.
  8. RajeshTrivedi Valued Senior Member

    The point is few here insisted that for the case in hand, the horizontal speed component can be ignored and it will not matter to the vertical motion, and a terminal speed in the vertical direction will be achieved even with non zero horizontal speed. My point was that terminal speed in this case cannot be achieved as long as horizontal speed component is present (#51).

    It is meaningless (and not correct) to compute the drag force by first splitting the velocity vector in two orthogonal directions (Horizontal , Vertical in this case) and then finding individual drag components in these directions by applying the formula on component velocities. The exact method is to first compute the drag force in the direction of motion and after that take the components in desired directions for further analysis. This method will conclude and clarify that horizontal speed in this case will impact the drag component in vertical direction, so as long as horizontal speed is non-zero the bullet will not achieve terminal speed.

    Let me compute based on your argument:

    1. Assume that at any instant the velocity vector for the bullet is 3i + 4j, and also assume that x = 1 (your argument, same for all). The resultant velocity is 5 m/seconds.

    A. Computing resultant drag force from the velocity components will give you

    Drag Force in x direction = xV^2 = 9
    Drag Force in y direction = xV^2 = 16

    Resultant Drag Force = Sqrt (81 + 256) = Sqrt(337) = 18.35 Units in the direction atan(81/256)

    B. Computing resultant drag force against the direction of motion method

    Resultant Drag Force = xV^2 = 25 Units in the direction atan(3/4).

    And the components shall be 25 X 3/5 = 15 Units and 25 X 4/5 = 20 Units.....not 9 units and 16 units as in A.

    B is the correct solution even if you take same 'x'. I do not know how you have arrived at same in your post #142, it cannot be. Probably you missed to square the force components to derive resultant force. (Did I not say that please do not get confused by presence of V^2 term in force formula)
  9. RajeshTrivedi Valued Senior Member

    So far so good.

    Not good. The sum 68x is not the resultant drag, the resultant drag is Sqrt(16x^2 + 4096x^2) = 64.12x along different direction.

    Hope you change your stand
  10. sculptor Valued Senior Member

    How do you factor in the destabilizing effects of the sonic transition?
    rate of spin of the bullet
    at what reduction of spin, will the bullet waffle or tumble?

    how will this effect velocity?
    Last edited: Feb 14, 2017
  11. Baldeee Valued Senior Member

    I'm not arguing against that.
    Please argue against what I have stated.
    Stop raising strawmen.
    Who said that that is what is being done.
    As I said, I was disputing your claim (or at least how I interpreted your claim) that the magnitude of drag when travelling at 824m/s (being 800/2oo in orthogonal directions) is not the same as the magnitude of drag when summing the individual drag calculations for 800 and 200.
    Everything else you typed here is a strawman.
    Stick to what I write, please.
    You don't know how I arrived at them being the same, even though the maths was set out for you, and so therefore you simply conclude that it cannot be??
    Not doing much to persuade people to take you seriously, are you.
    Again, and hopefully for the last time, asking someone not to get confused, and not actually doing anything to prevent that confusion in the first instance, makes the warning moot.

    But just to be clear: there was no confusion, just you seemingly unable to follow the maths.

    Let me show you again:
    Object travelling 8 m/s vertically and 2 m/s horizontally.
    Got it so far?

    The velocity is sqrt(68) m/s in the 2i+8j direction.
    Still with me?

    The drag for this object is proportional to velocity^squared, so is 68x.
    Got that?

    Now, the drag on the same object travelling 8m/s verticaly is....?
    That's right: 64x

    The drag on the same object travelling 2m/s horizontally is...?
    That's right: 4x

    What is the sum of those two?
    Drum roll, please....
    Ta da!
    That's right, it's 68x.

    Does or does that not match the 68x magnitude for the object travelling 8m/s vertically and 2m/s horizontally?

    And what's more it doesn't just work for 8 m/s and 2m/s, but is in fact a basic relationship of Pythagoras theorem.
    (Oops, there's that word you don't seem to like too much, I see).

    Now, before you get so arrogant as to just dismiss the result as clearly being due to some confusion you think I have... where is the error?
    You claim that the result "cannot be" and yet the maths is inescapable.
    Now show me the error or apologise.
    Or are you going to simply obfuscate your error away, that you didn't mean what you said, that I haven't understood it properly, that you're only referring to the OP, or some such nonsense?

    You do know that this is rather simple GCSE level maths, right?
    Here's a hint: what is the relationship between the magnitude of orthogonal sides to the diagonal in a right angled construction?
    Would it happen to be w^2 + l^2 = d^2 ? (Width, Lengh, Diagonal).
    Now what is the relationship I am having to explain to you about the magnitude of drag, expressed as the velocities?
    Would it happen to be Vh^2 + Vv^2 = Vd^2 ?

    Good golly, Miss Molly, I may be on to something here!
    Got it yet?
    Are you sure?

    I await your inevitable back-peddling.

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  12. Baldeee Valued Senior Member

    "Magnitude", for pete's sake.
    Read what is written!!
    As said, the drag is in the opposite direction to travel.
    This is about magnitude of that force.

    Got it yet?

    Needless to say, your inability to read does not serve you well.
  13. RajeshTrivedi Valued Senior Member


    You are making too much noise, I cannot match you in lung power.

    For a velocity vector 2i+8J, if the magnitude of drag force in -x direction is 4x and in -y direction it is 64x, then total is 68x? Thats what you are saying? Then it is wrong. Firstly 4x and 64x are meaningless quantities for drag force due to 2i+8j. The components of drag force in x and y direction, should be calculated from 68x (in the direction of -v). They will not be 4x and 64x, thus making these figures nonsensical.

    Secondly why be so selective about drag force? Why not momentum also? The magnitude (your choice of word) of momentum due to x component of velocity is 2m and due to y component of velocity is 8m, so your maths will give a total momentum of 10m! I am very sure your GCSE teacher did not teach you that, you prefer making noise rather than learning. I am sure now you can figure out the correct momentum value at least.

    Well I can understand your mistake, you are stuck due to presence of v^2 term in drag force calculation. Try some other power (instead of 2) and see if you are making any sense. And yes drag force can be modeled with non 2 powers, recall even the drag force is proportional to speed. another mistake you are making is jumping in without understanding my point of view. My point is very simple that drag force calculations can only be done on speed, not on the velocity components, thus velocity component in x direction impacts the total drag force and thus impacts terminal speed in y direction.
  14. Baldeee Valued Senior Member

    You can't seem to match where it matters, either.
    That is not what I am saying.
    I am saying that the magnitude of the drag force opposite the direction of travel (i.e. in the -2i-8j direction) is equal to the magnitude of the drag on 2i plus the magnitude of the drag on 8j.
    I have explained this in detail.
    I have provided the math to prove it is true.
    You see to want to create an argument where none should exist, so you seem to raise strawmen.
    Ever thought of sticking to what is written?
    Another strawman.
    Please stick to what is written.
    Or do you want me to argue about you claiming 2+2=5?
    And I am not selective about drag force.
    It holds true for any property that is the square of velocity, such as kinetic energy.
    Or are you now going to deny that the magnitude of the KE of the bullet travelling at 2i+8j is the sum of the magnitude of a bullet travelling at 2i plus the magnitude of a bullet travelling at 8j?
    You have simply worked yourself into knots about a very simple mathematical relationship and are clearly too stubborn to acknowledge it, feeling you have to argue against everything to try to recover.
    All you're doing now, though, is being a troll.
    There is no being stuck.
    Being proportional to V^2 is what makes what I am saying hold true, and it holds true for any property proportional to V^2.
    It is simple Pythagoras theorem.
    Ah, I see.
    You are now beginning to comprehend what a fool you have been arguing against what I have said so are merely compounding your strawmen with simply seeking to dismiss everything else.
    And yet no doubt you claim not to simply be a troll.
    So you now simply ignore all my criticisms of the point you made (and that I was specifically responding to) and simply return to a point you made to someone else?
    I guess trolls do as trolls do.

    Further, drag is only proportional to speed at low (i.e. non-turbulent) speeds, and even then it is only approximately proportional to speed.
    Heck, it can even be given as a constant in some scenarios.

    But in the bullet-through-air scenario, you are wrong in that drag force is proportional to the square of speed, not to speed as you claim above.
  15. RajeshTrivedi Valued Senior Member

    Too much of noise, and some conceptual mistakes.

    KE is a scalar quantity and also has quadratic (v^2) dependence, so it can be added, the way you are adding.
    But the force is not a scalar quantity. Your 4x and 64x are senseless figures, and even if they are accepted then directions will be -4xi and -64xj, so they cannot be added up to give you 68x in the direction of (-2i-8j).

    You are doing nothing but effectively using v^2 as (4+64), but goofing up by making it up as 4x and 64x as if they are some meaningful force values, here 4x and 64x are not the forces in -x and -y direction, thus making them useless for any further calculations. Like if you have to calculate the resultant force in vertical direction then it will be [mg - 68x * Sin(theta)] not (mg-64x).

    PS: Looking at the kind of noise you are making, I do not think you will accept, so I am off from this thread.
  16. Baldeee Valued Senior Member

    No conceptual mistakes from me, just blatant stubbornness and trolling from you.
    It's not a question of it being a scalar quantity or not, but of the mathematics involved in working out the solution.
    You seem to have missed that, or more likely your subsequent bluster is simply to obfuscate your failure.
    No more senseless than the numbers in any other addition you care to mention.
    And what is it about the term "magnitude" do you fail to comprehend?
    Magnitude is irrespective of direction.
    It is simply the size.
    Where have I said that they are forces in the x and y direction?
    That is simply your assumption from a failure to stick to what was actually written.
    So really just more strawmen from you, RT, to avoid concluding the inevitable.
    Given your obfuscation and strawmen, probably for the best, eh.
  17. origin In a democracy you deserve the leaders you elect. Valued Senior Member

    It is useless to try and have any meaningful discussion with RT, ignore is the best option...

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